2 Answers
2

In this particular case, if you scan down the right side of the truth table, you get almost all $1$'s. The sentence is false only if $p$ holds and $q$ fails. So the negation of our sentence is equivalent $p\land \lnot q$. It follows that our sentence is equivalent to $\lnot(p\land \lnot q)$, and therefore to $\lnot p\lor q$.

This kind of procedure is also useful when there are more proposition letters, but the sentence is "seldom" false (or seldom true).

It may have been done algebraically, rather than by means of a truth table.

Since $r\to s$ is logically equivalent to $\lnot r\lor s$, $p \wedge \neg q \to q \vee \neg p$ is logically equivalent to $\lnot(p\land\lnot q)\lor(q\lor\lnot p)$. Applying de Morgan’s law to the first term, we see that this in turn is logically equivalent to $\lnot p\lor q\lor q\lor\lnot p$. And of course $q\lor q$ and $\lnot p\lor\lnot p$ are logically equivalent to $q$ and to $\lnot p$, respectively, so $\lnot p\lor q\lor q\lor\lnot p$ reduces to $\lnot p\lor q$.